Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 42}{x + 10} = \dfrac{-19x - 132}{x + 10}$
Solution: Multiply both sides by $x + 10$ $ \dfrac{x^2 - 42}{x + 10} (x + 10) = \dfrac{-19x - 132}{x + 10} (x + 10)$ $ x^2 - 42 = -19x - 132$ Subtract $-19x - 132$ from both sides: $ x^2 - 42 - (-19x - 132) = -19x - 132 - (-19x - 132)$ $ x^2 - 42 + 19x + 132 = 0$ $ x^2 + 90 + 19x = 0$ Factor the expression: $ (x + 9)(x + 10) = 0$ Therefore $x = -9$ or $x = -10$ At $x = -10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -10$, it is an extraneous solution.